Petya sometimes has to water his field. To water the field, Petya needs a tank with exactly V ml of water.
Petya has got N tanks, i-th of them initially containing ai ml of water. The tanks are really large, any of them can contain any amount of water (no matter how large this amount is).
Also Petya has got a scoop that can contain up to K ml of water (initially the scoop is empty). This scoop can be used to get some water from some tank, and after that pour it all into some tank (it is impossible to get water from multiple tanks without pouring it, or leave some water in the scoop when pouring it). When Petya tries to get some water from a tank, he gets min(v, K) water, where v is the current volume of water in the tank.
Is it possible to obtain a tank with exactly V ml of water using these operations? If it is possible, print a sequence of operations that allows to do it. If there are multiple ways to obtain needed amount of water in some tank, print any of them.
Input
The first line contains 3 integers: N (2 ≤ N ≤ 5000), K (1 ≤ K ≤ 5000), and V (0 ≤ V ≤ 109) — the number of tanks, the maximum volume of water the scoop can contain, and the required amount of water in some tank, respectively.
The second line contains N integers ai (0 ≤ ai ≤ 105), where ai is initial volume of water in i-th tank.
Output
If it is impossible to obtain a tank with exactly V ml of water, print NO.
Otherwise print YES in the first line, and beginning from the second line, print the sequence of operations in the following format:
Each line has to contain 3 numbers denoting a compressed operation: “cnt x y” (1 ≤ cnt ≤ 109, 1 ≤ x, y ≤ N), where x is the index of the tank where we get water, y is the index of the tank where we pour water, and cnt is the number of times we transfer water from tank x to tank y.
The number of these lines must not exceed N + 5.
Examples
Input
2 3 5
2 3
Output
YES
1 2 1
Input
2 3 4
2 3
Output
NO
Input
5 2 0
1 3 5 7 9
Output
YES
2 2 1
3 3 1
4 4 1
5 5 1
Thinking Source: https: //www.cnblogs.com/kkkkahlua/p/8413054.html
meaning of the questions:
Do you have n ponds, there is a certain amount of water and the volume of infinity. K is a volume of a spoon, can hold a pool min (k, a [i] ) of the water into the other tank. Find out whether the distribution of water v volume.
Ideas:
firstly are converted to digital to count modulo k.
Define F. [I] [j] for the i-th front pond water dispenser feasibility of j.
F. [I] [J] =
0 - infeasible
1-- feasible, but not necessarily with a [i]
2 - feasible and must use a [i]
So with this we can see whether a volume v, and drawn from the results of reverse thrust must use what is and what is not necessarily want to use.
If the sum {a [i]]} or dp [n] [v% k] = 0, then certainly not.
If you do not have to use, then v must be a multiple of k, all the water together, add v / k spoon out of the water just fine
otherwise have to use all the water put together as s1, not We must use water together as s2. Then the gap s1 and v must be a multiple of k, plus (minus) the multiple fine.
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int a[5005],b[5005],dp[5005][5005];
int flag[5005];
int main()
{
int n,k,v;scanf("%d%d%d",&n,&k,&v);
int sum = 0;
for(int i = 1;i <= n;i++)
{
scanf("%d",&a[i]);
b[i] = a[i];a[i] %= k;
sum += b[i];
}
if(sum < v)
{
printf("NO\n");
return 0;
}
dp[0][0] = 1;
for(int i = 1;i <= n;i++)
{
for(int j = 0;j < k;j++)
{
if(dp[i - 1][j])
{
dp[i][j] = 1;
if(!dp[i][(j + a[i]) % k])dp[i][(j + a[i]) % k] = 2;
}
}
}
int num = v % k;
if(!dp[n][num])
{
printf("NO\n");
return 0;
}
printf("YES\n");
int x1 = -1,x2 = -1,ans = 0;
for(int i = n;i >= 1;i--)
{
if(dp[i][num] == 2)
{
flag[i] = true;
num = (num - a[i] + k) % k;
ans += b[i];
if(x1 == -1)x1 = i;
}
else if(x2 == -1)x2 = i;
}
if(x1 == -1)//没有一定要添加的，那就是添加任何一个都可以，那v一定是k的倍数
{
for(int i = 1;i < n;i++)
{
if(b[i])
{
printf("%d %d %d\n",(b[i] + k - 1) / k,i,n);
}
}
if(v / k)printf("%d %d %d\n",v / k,n,1);
return 0;
}
int res = v - ans;
int s1 = b[x1],s2 = b[x2];
for(int i = 1;i <= n;i++)
{
if(i == x1 || i == x2 || !b[i])continue;
if(!flag[i])
{
printf("%d %d %d\n",(b[i] + k - 1) / k,i,x2);//非必要的全部放到x2
s2 += b[i];
}
else
{
printf("%d %d %d\n",(b[i] + k - 1) / k,i,x1);//必要的全部放到x1
s1 += b[i];
}
}
int cnt = (v - s1) / k;
if(cnt > 0)printf("%d %d %d\n",cnt,x2,x1);//此时x1与v的差一定是k的倍数
else if(cnt < 0)printf("%d %d %d\n",-cnt,x1,x1 == 1 ? 2 : 1);
return 0;
}
